JEE Main 2025 Apr 4 Shift 1, Physics Q25: Motional EMF

The motional EMF along a curved/bent wire moving through a uniform magnetic field depends only on the straight-line displacement vector between the endpoints, not on the actual path. So replace the zig-zag ABCDE with a straight conductor from A to E.

From the geometry (each 10 cm arm at 90°), the straight-line distance between A and E (component perpendicular to $\vec{v}$) is ${\ell }_{AE}=2\cdot (10\sin 45^{\circ })=10\sqrt{2}$ cm $=0.10\sqrt{2}$ m.

EMF: $\epsilon =Bv{\ell }_{AE}=\frac{1}{\sqrt{2}}\cdot 0.10\cdot 0.10\sqrt{2}=0.01$ V $=10$ mV.